Let $y=\dfrac{4x^2+3x}{2x-7}$. $\dfrac{dy}{dx}=$
Answer: $\dfrac{4x^2+3x}{2x-7}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d y d x = d d x ( 4 x 2 + 3 x 2 x − 7 ) = ( 2 x − 7 ) d d x ( 4 x 2 + 3 x ) − ( 4 x 2 + 3 x ) d d x ( 2 x − 7 ) ( 2 x − 7 ) 2 The quotient rule = ( 2 x − 7 ) ( 8 x + 3 ) − ( 4 x 2 + 3 x ) ( 2 ) ( 2 x − 7 ) 2 Differentiate ( 4 x 2 + 3 x ) & ( 2 x − 7 ) = 16 x 2 + 6 x − 56 x − 21 − 8 x 2 − 6 x ( 2 x − 7 ) 2 Expand = 8 x 2 − 56 x − 21 ( 2 x − 7 ) 2 \begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{4x^2+3x}{2x-7}\right) \\\\ &=\dfrac{(2x-7)\dfrac{d}{dx}(4x^2+3x)-(4x^2+3x)\dfrac{d}{dx}(2x-7)}{(2x-7)^2} \gray{\text{The quotient rule}} \\\\ &=\dfrac{(2x-7)(8x+3)-(4x^2+3x)(2)}{(2x-7)^2} \gray{\text{Differentiate }(4x^2+3x)\text{ & }(2x-7)} \\\\ &=\dfrac{16x^2+6x-56x-21-8x^2-6x}{(2x-7)^2} \gray{\text{Expand}} \\\\ &=\dfrac{8x^2-56x-21}{(2x-7)^2} \end{aligned} In conclusion, $\dfrac{dy}{dx}=\dfrac{8x^2-56x-21}{(2x-7)^2}$, or any other equivalent form.